Question

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Q1: What will be the pressure (in mmHg) inside of a 50.0 L container that holds 7.00 moles of hydrogen gas at 0.0 C?

Q2: Oxygen Gass is produced for use in a small-scale experiment by the catalytic decomposition of hydrogen peroxide:

2 H2O2 (aq) -------------------> 2 H2O (l) + O2 (g)

If 50.0 mL of a 1.00 M solution of H2O2 completely decomposes and what volume of dry oxygen gas can be collected at 21.5 C and 742 mmHg?

Answer 1

Please show how work on how you got this answer.

Q1: What will be the pressure (in mmHg) inside of a 50.0 L container that holds 7.00 moles of hydrogen gas at 0.0 C

Given: Volume of container = 50.0 L , mole of Hydrogen gas = 7.00

Temperature = 0.0 deg C = 0 deg C + 273.15 = 273.15 K

We use ideal gas law to find out pressure

PV= nRT

P = pressure in atm

V = volume in L

n = number of moles

T = temperature in K

R = gas constant = 0.08206 L atm / ( mol K )

Lets plug all values in above equation to get pressure

P = nRT / V

= 7.00 mol * 0.08206 L atm /( K mol ) * 273.15 K ) / 50.0 L

= 3.14 atm

1 atm = 760 mmHg

We use this relation to get pressure in mmHg

= 3.14 atm * 760 mmHg / 1 atm
= 2384.92 mmHg

Q2: Oxygen Gass is produced for use in a small-scale experiment by the catalytic decomposition of hydrogen peroxide:

2 H2O2 (aq) -------------------> 2 H2O (l) + O2 (g)

If 50.0 mL of a 1.00 M solution of H2O2 completely decomposes and what volume of dry oxygen gas can be collected at 21.5 C and 742 mmHg?

Given : Volume of H2O2 = 50.0 mL , Molarity = 1.00 M

Temperature = 21.5 deg C = 21.5 deg C + 273.15 =294.65 K

Pressure = 742 mmHg

Solution :

Conversion of pressure to atm

= 742 mmHg * 1 atm/ 760 mmHg

= 0.98 atm

We have to find volume of O2

By using reaction stoichiometry we find the moles of O2

Mol of H2O2 = Volume in L * molarity = 0.050 L * 1.0M = 0.050 mol

Calculation of moles of O2 from moles of H2O2

Moles of O2 = moles of H2O2 * 1 mol O2 / 2 mol H2O2

= 0.050 mol H2O2 * 1 mol O2 / 2mol H2O2

= 0.025 mol O2

Ideal gas law

PV = nRT

V = nRT / P

= 0.025 mol * 0.08206 L atm / ( K mol ) * 294.65 K ) / 0.98 atm

= 0.619 L

= 619.0 mL

Volume of O2 collected = 619.0 mL